$$ 1 \leftarrow \begin{array}{l} ext { Researchers conducted an experiment to test the effects of alcohol. Erors were recorded in a test of visual and motor skills for a treatment group of } 29 ext { people who drank ethanol and another group of } 29 ext { people given a } \ ext { placebo. The errors for the treatment group have a standard deviation of } 230 ext {, and the errors for the placebo group have a standard deviation of } 0.87 ext {. Assume that the two populations are normally distributed. Use a } 0.05 \ ext { significance level to test the claim that both groups have the same amount of variation among the errors. }\end{array} $$

Question

$$ 1 \leftarrow \begin{array}{l}	ext { Researchers conducted an experiment to test the effects of alcohol. Erors were recorded in a test of visual and motor skills for a treatment group of } 29 	ext { people who drank ethanol and another group of } 29 	ext { people given a } \ 	ext { placebo. The errors for the treatment group have a standard deviation of } 230 	ext {, and the errors for the placebo group have a standard deviation of } 0.87 	ext {. Assume that the two populations are normally distributed. Use a } 0.05 \ 	ext { significance level to test the claim that both groups have the same amount of variation among the errors. }\end{array} $$

UpStudy AI · Accepted Answer

**Step 1. Define the hypotheses**

We wish to test whether the variances are equal. Let  
$$
\sigma_1^2 \text{ be the variance of the treatment group, and } \sigma_2^2 \text{ be the variance of the placebo group.}
$$
Then the hypotheses are:
- Null hypothesis: $$ H_0: \sigma_1^2 = \sigma_2^2 $$
- Alternative hypothesis: $$ H_a: \sigma_1^2 \neq \sigma_2^2 $$

---

**Step 2. Compute the sample variances**

For the treatment group we have:
- Sample size: $$ n_1 = 29 $$
- Sample standard deviation: $$ s_1 = 230 $$
  
Thus, the sample variance is:
$$
s_1^2 = 230^2 = 52900.
$$

For the placebo group we have:
- Sample size: $$ n_2 = 29 $$
- Sample standard deviation: $$ s_2 = 0.87 $$
  
Thus, the sample variance is:
$$
s_2^2 = 0.87^2 \approx 0.7569.
$$

---

**Step 3. Determine the test statistic**

For testing the equality of two variances with normal populations, we use the $$ F $$-statistic given by:
$$
F = \frac{s_1^2}{s_2^2}.
$$
Choosing the larger sample variance as the numerator (here, $$ 52900 > 0.7569 $$), we have:
$$
F = \frac{52900}{0.7569} \approx 69895.
$$

---

**Step 4. Degrees of freedom**

Since both samples have $$ n = 29 $$, the degrees of freedom are:
- Numerator: $$ df_1 = n_1 - 1 = 28 $$,
- Denominator: $$ df_2 = n_2 - 1 = 28 $$.

---

**Step 5. Decision rule**

At a significance level of $$\alpha = 0.05$$ for a two-tailed test, we need the lower and upper critical values:
- The lower critical value is $$ F_{0.025}(28,28) $$,
- The upper critical value is $$ F_{0.975}(28,28) $$.

Because the computed test statistic $$ F \approx 69895 $$ is extremely large and far exceeds the typical upper critical value (which would be a number not anywhere near 69895), we reject $$ H_0 $$.

---

**Step 6. Conclusion**

Since the test statistic is in the extreme tail of the $$ F $$-distribution and the $$ p $$-value will be extremely small, we have sufficient evidence at the $$ \alpha = 0.05 $$ level to reject the null hypothesis.

Thus, we conclude that there is significant evidence to suggest that the two groups do not have the same amount of variation among the errors.
There is significant evidence to suggest that the treatment and placebo groups have different amounts of variation in their errors.

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Statistics Questions